\newproblem{lay:5_1_26}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 5.1.26}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Show that if $A^2$ is the zero matrix, then the only eigenvalue of $A$ is 0.
}{
  % Solution
	Suppose $\lambda$ is an eigenvalue of $A$
	\begin{center}
		$A\mathbf{x}=\lambda\mathbf{x}$
	\end{center}
	Let's multiply on both sides by $A$
	\begin{center}
		$A^2\mathbf{x}=\lambda A\mathbf{x}$ \\
		$\mathbf{0}=\lambda (\lambda \mathbf{x})$ \\
		$\mathbf{0}=\lambda^2\mathbf{x}$ \\
	\end{center}
	But $\mathbf{x}$ is non-zero (by the definition of eigenvector). Then, $\lambda^2=0$ and this implies that the only eigenvalue of $A$ is 0.
}
\useproblem{lay:5_1_26}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
